10. Odd, Even, Pos, Neg Flashcards

1
Q

What are integers?

A

Positive (1, 2, 3, etc.) or negative (-1, -2, -3, etc.) whole numbers that have no fractional part. Integers include counting numbers, their negative counterparts and zero

  • Adding, subtracting and multiplying integers always give you an integer
  • Dividing an integer by another integer results in an integer or a fraction
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2
Q

What is an even number?

A

Integers that are divisible by 2. 14 is even because 14/2 = integer (7)

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3
Q

What is an odd number?

A

Integers that are not divisible by 2. 15 is not even because 15/2 does not equal an integer (7.5)

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4
Q

What are the addition/subtraction rules for odds and evens?

A
  • Add or subtract 2 odds or 2 evens, and the result is EVEN
  • Add or subtract an odd with an even, and the result is ODD

Even ± Even = Even (e.g. 2 + 2 = 4)
Odd ± Odd = Even (e.g. 3 + 3 = 6
Even ± Odd = Odd (e.g. 2 + 3 = 5)

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5
Q

What are the multiplication rules for odds and evens?

A
  • When you multiply integers, if ANY of the integers is even, the result is EVEN
  • Likewise, if NONE of the integers is even, then the result is ODD

Even * Even = Even (e.g. 2 * 2 = 4) ALSO divisible by 4
Even * Odd = Even (e.g. 2 * 3 = 6)
Odd * Odd = Odd (e.g. 3 * 3 = 9)

*NOTE: if you multiply together several even integers, the result will be divisible by higher and higher powers of 2 (e.g. 2 * 5 * 6 * 10 = 600 which is divisible by 8 = 2^3)

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6
Q

What are the division rules for odds and evens?

A

Even / Even = Anything (even, odd or not an integer)
Even / Odd = Even or not an integer
Odd / Even = Not an integer
Odd / Odd = Odd or not an integer

*NOTE: an odd integer divided by any other integer CANNOT produce an even integer

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7
Q

What do you know about (prime + prime) = odd?

A
  • One of those primes must be the number 2

- Conversely, if you know that 2 CANNOT be one of the primes in the sum, then the sum of the two primes must be even

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8
Q

If a and b are both prime numbers greater than 10, which of the following CANNOT be true?
I. ab is an even number
II. the difference between a and b = 117
III. the sum of a and b is even

A

I and II Only. a and b must be odd

I. ab must be an odd number (CANNOT be true)
II. odd – odd = even (CANNOT be true)
III. odd + odd = even (TRUE)

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9
Q

If x > 1, what is the value of integer x?

(1) There are x unique factors of x
(2) The sum of x and any prime number larger than x is odd

A

Answer A.

(1) SUFFICIENT. In order for this to be true, every integer between 1 and x, inclusive, must be a factor of x. Testing numbers you can see that this property holds for 1 and for 2 only. Therefore, this x = 1 or 2. However, the original problem says that x > 1, so x must equal 2
(2) INSUFFICIENT. x must equal at least 2, so this includes prime numbers larger than 2. Therefore, the prime number is odd, and x is even. However, this does not tell you which even number x could be

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10
Q

What is the remainder when a is divided by 4?

(1) a is the square of an odd integer
(2) a is a multiple of 3

A

Answer A. Even numbers are multiples of 2, so any arbitrary even number can be written as 2n, where n is any integer. Odd numbers are one more or less than multiples of 2, so an arbitrary odd number can be written as 2n + 1 or 2n – 1, where n is an integer

(1) SUFFICIENT. The square of an arbitrarily odd number can be written as (2n + 1)^2 = 4n^2 + 4n + 1. The first two terms of this expression are multiples of 4, which have a remainder of 0 when divided by 4. The third term, 1, gives a remainder of 1 when divided by 4.
(2) INSUFFICIENT. Test numbers. When 3 is divided by 4, the remainder is 3. When 6 is divided by 4, the remainder is 2.

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11
Q

If a, b, and c are integers and ab + c is odd, which of the following must be true?
I. a + c is odd
II. b + c is odd
III. abc is even

A

III Only. A table is a good tool for keeping track of different scenarios. Two steps: (1) ask yourself, what would need to be true in order for ab + c to be odd? An odd plus and even equals an odd, so either ab or c needs to be even (2) write out only those scenarios in which either ab or c is even

(1) a = ODD, b = ODD, c = EVEN, ab + c = ODD
(2) a = ODD, b = EVEN, c = ODD, ab + c = ODD
(3) a = EVEN, b = ODD, c = ODD, ab + c = ODD
(4) a = EVEN, b = EVEN, c = ODD, ab + c = ODD

I. a + c is odd: Scenario (2) goes against this statement
II. b + c id odd: Scenario (3) goes against this statement
III. abc is even: For abc to be even, only one of the three variables needs to be even. In all four scenarios, at least one of the variables is even. MUST BE TRUE

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12
Q

How should you deal with the scenarios in an odds and evens problem?

A

(1) Ask yourself: how do any given constraints limit the possible scenarios?
(2) Create a table to keep track of the allowable scenarios, given any constraints in the problem

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13
Q

Number Properties Guide, Ch 2, Q 9. If x, y, and z are prime numbers and x < y < z, what is the value of x?

(1) xy is even
(2) xz is even

A

Answer D.

(1) SUFFICIENT. If xy is even, then x is even or y is even. Since x < y, x must be 2, because 2 is the smallest and only even prime number
(2) SUFFICIENT. Similarly, if xz is even, then x is even or z is even. Since x < z, x must equal 2, because 2 is the smallest and only even prime number

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14
Q

Number Properties Guide, Ch 5, Q 1. If x, y and z are integers, is x even?

(1) 10^x = (4^y)(5^z)
(2) 3^(x + 5) = 27^(y + 1)

A

Answer A. Odds & Evens.

(1) SUFFICIENT. You can break the bases down into prime factors: (2^x)(5^x) = (2^2y)(5^z). This means that x = 2y and x = z. y is an integer, so x must be even, because x = 2y
(2) INSUFFICIENT. You can again break the bases down into prime factors. 3^(x + 5) = 3^(3y + 3). This tells you that x + 5 = 3y + 3, so x + 2 = 3y. y is an integer, so x must be 2 larger than a multiple of 3, but that does not tell you whether x is even. If y = 1, then x = 1 (odd), but if y = 2, then x = 4 (even)

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15
Q

Number Properties Guide, Ch 5, Q 3. If c and d are integers, is c – 3d even?

(1) c and d are odd
(2) c – 2d is odd

A

Answer A. Odds & Evens.

(1) SUFFICIENT. If both c and d are odd, then c – 3d = Odd – (3 * Odd) = Odd – Odd = Even
(2) INSUFFICIENT. If c – 2d is odd, then c must be odd, because 2d will always be even. However, this tells you nothing about d

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16
Q

Number Properties Guide, Ch 5, Q 6. Is the integer x odd?

(1) 2(y + x) is an odd integer
(2) 2y is an odd integer

A

Answer E. Odd & Evens.

(1) INSUFFICIENT. 2(y + x) is an odd integer. How is it possible that 2 multiplied by something could yield an odd integer? The value in parenthesis must not be an integer itself. List some possibilities:
2(y + x) = 1, 3, 5, 7, 9, etc.
(y + x) = 1/2, 3/2, 5/2, 7/2, 9/2, etc.
You know that x is an integer, so y must be a fraction in order to get such a fractional sum. Say that y = 1/2. In that case, x = 0, 1, 2, 3, 4, etc. Thus, x can be either odd (YES) or even (NO)

(2) INSUFFICIENT. This statement tells you nothing about x. If 2y is an odd integer, this implies that y = odd/2 = 1/2, 3/2, 5/2, etc.
(C) INSUFFICIENT. Statement (2) fails to eliminate the case you used in Statement (1) to determine that x can be either odd or even. Thus, you still cannot answer the question with a definite yes or no.

17
Q

What are positive numbers?

A

-To the right of the zero on a number line

NOTE: Zero itself is neither positive nor negative.

18
Q

What are negative numbers?

A

-To the left of the zero on a number line

NOTE: Zero itself is neither positive nor negative.

19
Q

What does the absolute value of a number tell you?

A

How far away the number is from 0 on the number line

Remember: The absolute value is ABSOLUTELY POSITIVE! (except for 0 because ǀ0ǀ = 0, which is the smallest possible absolute value)

20
Q

What are the rules of multiplication and division for positive and negative numbers?

A
  • If signs are the same, the answer is positive, but if not, the answer is negative
  • When you multiply or divide a group of non-zero numbers, the result will be negative if you have an odd number of negative numbers

Positive * Positive = Positive
Positive * Negative = Negative
Negative * Negative = Positive

21
Q

Is the product of all of the elements in Set S negative?

(1) All of the elements in Set S are negative
(2) There are 5 negative number in Set S

A

Answer C.

(1) INSUFFICIENT. We do not know how many negative numbers there are. If there are an odd number of negatives, then the result is negative. If there are an even number of negatives, then the result is positive
(2) INSUFFICIENT. The statement does not tell us whether or not there are other numbers in the set. For instance, there could be 5 negative numbers as well as a few other numbers. If zero is one of those numbers, then the product will be zero, and zero is not negative
(C) SUFFICIENT. Combined, you know that Set S contains 5 negative numbers and nothing else. The product of elements in Set S must be negative.

22
Q

When does the order of numbers matter in an equation?

A

*Order does not matter: addition and multiplication
2 x 3 = 3 x 2 = 6
2 + 3 = 3 + 2 = 5

*Order matters: subtraction and division
3 – 2 does not equal 2 – 3
4 / 2 does not equal 2 / 4

23
Q

What happens when you divide by zero or divide zero by another number?

A

Divide by zero = undefined

Zero divided by another number = 0

24
Q

Is zero consider even or odd, positive or negative?

A

Zero is even. 0/2 = 0. The result is an integer, so the number 0 is divisible by 2. As a result, the number 0 is even.

Zero is neither positive nor negative

25
Q

If (a – b)/c < 0, is a > b?

(1) c < 0
(2) a + b < 0

A

Answer A. This problem is a disguised positives & negatives question. Generally speaking, whenever you see inequalities with zero on either side of the inequality, you should consider testing positive/negative cases to help solve the problem

(1) SUFFICIENT. c is negative, thus a – b must be positive. a – b > 0 or a > b
(2) INSUFFICIENT. This does not tell us whether a is larger than b

26
Q
If ab > 0, which of the following must be negative?
A. a + b
B. ǀaǀ + b 
C. b – a
D. a/b
E. –a/b
A

Answer E. Set up a scenario chart to help evaluate the problem

(1) a = pos, b = pos, ab > 0? YES
(2) a = pos, b = neg, ab > 0? NO
(3) a = neg, b = pos, ab > 0? NO
(4) a = neg, b = neg, ab > 0? YES

Only two scenarios match. Either both a and b are positive or they are both negative. Going through the answers choices we can determine that E is the only option that must be negative. If a and b are both positive, -a/b will be negative. If a and b are both negative, -a/b will still be negative